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Centripetal Force: The center-seeking force
Group Members: Joshua Velez, Patrick Hannigan-Devine, and Eric Guidarelli
PURPOSE:
To move in a circle F=ma is required, where acceleration being the rate of change of velocity with velocity being both magnitude and direction. Magnitude of acceleration can be found by a=v2R. The relation of these two is found in centripetal force F=mv2R. This lab will prove the relation of the first two equations.
EQUIPMENT:
* Centripetal Force Apparatus * Mass Balance * Mass Set * Eye Protection * Mass Hanger * Level * Stopwatch * Ruler

Draw two Force Diagrams: Draw separate diagrams for the bob and for the hanging mass

Fg
Fg
FT
FT
Fg
Fg
FSpring
FSpring
FT
FT
FT
FT

QUESTIONS WITHIN PROCEDURES:
Why is it important for the string and the spring to be horizontal and collinear? It is important so that there are no components of either force acting on the bob. This allows us to relate the weight of the hanging mass directly to the force of the spring.
Is there any change in the force of the spring if the bob’s mass changes?
A change in the bob’s mass will not affect the force of the spring because the bob’s weight is a force that acts only along the y-axis. This is another reason why it is important to have the string and the spring horizontal and collinear.
How does the force change when the mass of the bob changes?
It doesn’t.
How has the velocity changed? If the mass is larger the velocity is slower and if the mass is smaller, then the velocity is faster. How does the centripetal force change with radius?
The centripetal force is greater with a larger radius and the force is smaller with a smaller radius.
DATA:
Table One: Calibration of the Spring Force Experimental Data | Calculations | Mass ofBob (kg) | Radius (Lengthfrom center of Bob to center of Rod) (m) | Hanging masses, mH(kg) | Hanging Weight, mHg(N) | Force of Spring for this Radius. | .3594 | 14 | .55 | 5.39 | 5.39 | .4594 | 14 | .55 | 5.39 | 5.39 | .3594 | 17 | 1.09 | 10.682 | 10.682 |

Table Two: Measurement of period, calculation of required centripetal force Bob Mass,mB | Radius,R | Trial No. | N | Total time | PeriodΤ | Ave.Τ | v=2πRΤ | F=mv2R | Fspring (from above) | % Diff. | .3594 | .14 | 1 | 50 | 31.2 | .624 | .6236 | 1.411 | 5.111 | 5.39 | 5.18 | | | 2 | 50 | 31.5 | .63 | | | | | | | | 3 | 50 | 30.85 | .617 | | | | | | .4594 | .14 | 1 | 50 | 32.67 | .6534 | .6531 | 1.347 | 5.954 | 5.39 | 10.46 | | | 2 | 50 | 32.7 | .654 | | | | | | | | 3 | 50 | 32.6 | .652 | | | | | | .3594 | .17 | 1 | 50 | 25.67 | .5134 | .5019 | 2.128 | 9.574 | 10.682 | 10.37 | | | 2 | 50 | 24.58 | .4916 | | | | | | | | 3 | 50 | 25.03 | .5006 | | | | | |

Questions
1. If it is suddenly removed, what would happen to the Bob?
The bob would continue moving in the direction of its instantaneous velocity at the moment the string (the force) was removed.
2. What does Newton's First Law predict would happen to the bob if all force were removed?
Is this consistent with your answer to Question 1?
Newton’s law would allow one to predict that once the force on the bob was removed, the bob would have no acceleration and its velocity therefore would remain constant. This is consistent with our answer to question one.
3. Examine equation (3). If everything in the equation were constant except that the mass increased, how would the centripetal force required to move in a circle change?
If you double the mass what happens to the required centripetal force?
From equation three in the lab manual, if everything except the mass remained constant and the mass were increased, the centripetal force required to move in a circle would also increase. If you double the mass, the required centripetal force is also doubled.
4. Examine equation (3). If everything in the equation were constant except that the speed increased, how would the centripetal force required to move in a circle change? If you double the speed what happens to the required centripetal force?
From equation three in the lab manual, if everything except the velocity remained constant and the velocity were increased, the centripetal force required to move in a circle would also increase. If you double the velocity, the required centripetal force would be quadrupled.
5. In your experiment, you changed the mass but kept the radius constant. Because the radius did not change, the force of the spring did not change. Instead, when you increased the mass, you had to change something else in order to make the required centripetal force stay the same. What did you change? Did you increase or decrease it? By how much (what ratio)? Is this what you would have predicted from equation (3)?
In order for the required centripetal force to stay the same, the period had to increase. The period increased by 0.0295 seconds. This is a ratio of 1.0473 times the first period. This is consistent with what we predicted from equation three because as the period increases, the velocity decreases and within the equation if the mass is increased and the velocity is decreased, both by the right ratio, then the centripetal force remains the same.
6. When you increased the radius of motion the force applied by the stretched spring increased.
How did the velocity for circular motion change?
When we increased the radius of motion, the velocity for circular motion also increased. This agrees with the equation that v = 2*pi*R/T

CONCLUSION:
Students successfully understood, learned from, and applied the equations discussed in the lab. They were able to calculate the centripetal acceleration required to keep a mass in constant circular motion and also were able to calculate an average velocity from measuring the period of a bob in circular motion. Students were also able to apply Newton’s laws throughout the lab and make great use of F = ma within their calculations and predictions.